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A Point source of light B is placed at a distance L in front of centre of a mirror of width d hung vertically on a wall. A man walks in front of the mirror along a line parallel to mirror at a distance 2L from it as shown. Find greatest distance over which he can see the image of light source in mirror.

$(a)\;d \\ (b)\;3d \\ (c)\;2d \\ (d)\;4d $

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The observer can see through distance P'Q', where reflected ray from B can meet line.
From geometry $CD=d=c'P'$
$P' Q'=C'D'+P'D'+Q'C'$
$\qquad= d+2PD+2CQ$
$\qquad= d+ 2( PD+qC)$
$\qquad= d+2d$
Hence b is the correct answer.
answered Feb 10, 2014 by meena.p
edited Mar 8 by sharmaaparna1

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